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arXiv:math/0611300v1 [math.NT] 10 Nov 2006 RAMANUJAN’S IDENTITIES AND REPRESENTATION OF INTEGERS BY CERTAIN BINARY AND QUATERNARY QUADRATIC FORMS ALEXANDER BERKOVICH AND HAMZA YESILYURT Abstract. We revisit old conjectures of Fermat and Euler regarding representation of integers by binary quadratic form x 2 +5y 2 . Making use of Ramanujan’s 1 ψ 1 summation formula we establish a new Lambert series identity for P n,m=-∞ q n 2 +5m 2 . Conjectures of Fermat and Euler are shown to follow easily from this new formula. But we don’t stop there. Employing various formulas found in Ramanujan’s notebooks and using a bit of ingenuity we obtain a collection of new Lambert series for certain inﬁnite products associated with quadratic forms such as x 2 +6y 2 ,2x 2 +3y 2 , x 2 + 15y 2 , 3x 2 +5y 2 , x 2 + 27y 2 , x 2 + 5(y 2 + z 2 + w 2 ), 5x 2 + y 2 + z 2 + w 2 . 1. Introduction A binary quadratic form (BQF) is a function Q(x,y)= ax 2 + bxy + cy 2 with a,b,c Z. It will be denoted by (a,b,c). We say that n is represented by (a,b,c) if there exist x and y Z such that Q(x,y)= n. The representation theory of BQF has a long history that goes back to antiquity. Diophantus’ Arithmeticae contains the following important example of composition of two forms (x 2 1 + y 2 1 )(x 2 2 + y 2 2 )=(x 1 x 2 y 1 y 2 ) 2 +(x 1 y 2 + x 2 y 1 ) 2 . Inﬂuenced by Diophantus, Fermat studied representations by (1, 0,a). For a =1, 2, 3 he proved a number of important results such as the following. A prime p can be written as a sum of two squares iﬀ p 1 (mod 4). We remark that representation by (1, 0, 3) played an important role in Euler’s proof of Fermat’s Last Theorem in the case of n = 3. Fermat realized that (1, 0, 5) was very diﬀerent from the previous cases (1, 0, 1), (1, 0, 2) and (1, 0, 3) considered by him. He made the following conjecture. If p and q are two primes that are congruent to 3 or 7 (mod 20), then pq is representable by (1, 0, 5). Euler made two conjectures that were very similar to those of Fermat: a. Prime p is represented by (1, 0, 5) iﬀ p 1 or 9 (mod 20). b. If p is prime then 2p is represented by (1, 0, 5) iﬀ p 3 or 7 (mod 20). However, his next conjecture for (1, 0, 27) contained an unexpected cubic residue condition: Prime p is represented by (1, 0, 27) iﬀ p 1 (mod 3) and 2 is a cubic residue modulo p. Lagrange and Legendre initiated systematic study of quadratic forms. But it was Gauss who brought the theory of BQF to essentially its modern state. He introduced class form groups and genus theory for BQF. He proved Euler’s conjecture for (1, 0, 27) and in the process discovered a so-called cubic reciprocity law. Gauss’ work makes it clear why (1, 0, 27) is much harder to deal with than (1, 0, 5). Indeed, a class form group with discriminant 20 consist of two inequivalent classes (1, 0, 5) and (2, 2, 3). These forms can’t represent the same integer. On the other hand, a class form group with discriminant 108 consist of three classes (1, 0, 27), (4, 2, 7), (4, 2, 7). These forms belong to 2000 Mathematics Subject Classiﬁcation. Primary: 11E16, 11E25, 11F27, 11F30; Secondary: 05A19, 05A30, 11R29. Key words and phrases. quadratic forms, q-series identities, eta-quotients, multiplicative functions. 1
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### Transcript of arxiv.org › pdf › math › 0611300v1.pdf · 2019-03-21 · arXiv:math/0611300v1 [math.NT] 10...

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0611

300v

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10

Nov

200

6

RAMANUJAN’S IDENTITIES AND REPRESENTATION OF INTEGERS BY

CERTAIN BINARY AND QUATERNARY QUADRATIC FORMS

ALEXANDER BERKOVICH AND HAMZA YESILYURT

Abstract. We revisit old conjectures of Fermat and Euler regarding representation of integers bybinary quadratic form x2 + 5y2. Making use of Ramanujan’s 1ψ1 summation formula we establish

a new Lambert series identity forP

n,m=−∞qn

2+5m

2. Conjectures of Fermat and Euler are shown

to follow easily from this new formula. But we don’t stop there. Employing various formulas foundin Ramanujan’s notebooks and using a bit of ingenuity we obtain a collection of new Lambert seriesfor certain infinite products associated with quadratic forms such as x2 +6y2, 2x2 +3y2, x2 +15y2,3x2 + 5y2, x2 + 27y2, x2 + 5(y2 + z2 + w2), 5x2 + y2 + z2 + w2.

1. Introduction

A binary quadratic form (BQF) is a function

Q(x, y) = ax2 + bxy + cy2

with a, b, c ∈ Z. It will be denoted by (a, b, c). We say that n is represented by (a, b, c) if there existx and y ∈ Z such that Q(x, y) = n.

The representation theory of BQF has a long history that goes back to antiquity. Diophantus’Arithmeticae contains the following important example of composition of two forms

(x21 + y21)(x22 + y22) = (x1x2 − y1y2)

2 + (x1y2 + x2y1)2.

Influenced by Diophantus, Fermat studied representations by (1, 0, a). For a = 1, 2, 3 he proved anumber of important results such as the following.

A prime p can be written as a sum of two squares iff p ≡ 1 (mod4).We remark that representation by (1, 0, 3) played an important role in Euler’s proof of Fermat’s

Last Theorem in the case of n = 3.Fermat realized that (1, 0, 5) was very different from the previous cases (1, 0, 1), (1, 0, 2) and (1, 0, 3)

considered by him. He made the following conjecture.If p and q are two primes that are congruent to 3 or 7 (mod 20), then pq is representable by (1, 0, 5).Euler made two conjectures that were very similar to those of Fermat:a. Prime p is represented by (1, 0, 5) iff p ≡ 1 or 9 (mod 20).b. If p is prime then 2p is represented by (1, 0, 5) iff p ≡ 3 or 7 (mod 20).However, his next conjecture for (1, 0, 27) contained an unexpected cubic residue condition:Prime p is represented by (1, 0, 27) iff p ≡ 1 (mod3) and 2 is a cubic residue modulo p.Lagrange and Legendre initiated systematic study of quadratic forms. But it was Gauss who

brought the theory of BQF to essentially its modern state. He introduced class form groups and genustheory for BQF. He proved Euler’s conjecture for (1, 0, 27) and in the process discovered a so-calledcubic reciprocity law. Gauss’ work makes it clear why (1, 0, 27) is much harder to deal with than(1, 0, 5). Indeed, a class form group with discriminant −20 consist of two inequivalent classes (1, 0, 5)and (2, 2, 3). These forms can’t represent the same integer. On the other hand, a class form groupwith discriminant −108 consist of three classes (1, 0, 27), (4, 2, 7), (4,−2, 7). These forms belong to

2000 Mathematics Subject Classification. Primary: 11E16, 11E25, 11F27, 11F30; Secondary: 05A19, 05A30, 11R29.Key words and phrases. quadratic forms, q-series identities, eta-quotients, multiplicative functions.

1

2 ALEXANDER BERKOVICH AND HAMZA YESILYURT

the same genus. That is, they may represent the same integer. An interested reader may want toconsult [9] and [14] for the wealth of historical information and [19] for the latest development.

In his recent book, Number Theory in the Spirit of Ramanujan, Bruce Berndt discusses represen-tation problem for (1, 0, 1), (1, 0, 2), (1, 1, 1), (1, 0, 3). Central to this approach is Ramanujan’s 1ψ1

summation formula which implies in particular that [7, p.58, eq. (3.2 .90)]

(1.1)∑

x,y∈Z

qx2+y2

= 1 + 4∑

n≥1

qn

1 + q2n.

Using geometric series it is straightforward to write the right hand side of (1.1) as

1 + 4∑

n≥1,m≥0

(−1)mqnq2nm = 1 + 4∑

n≥1,m≥0

(−1)mqnqn(2m+1)

= 1 + 4∑

n≥1,m≥1

(−4

m

)qnm

= 1 + 4∑

n≥1

d|n

(−4

d

)qn,

where we used the Kronecker symbol to be defined in the next section and the well known formula(−4

n

)=

{0 if n is even,

(−1)(n−1)/2 if n is odd.

Let r(n) be the number of representations of a positive integer n by quadratic form k2+ l2. Supposethe prime factorization of n is given by

n = 2ar∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1 (mod4) and qi ≡ −1 (mod4).

Using the fact that∑

d|n

(−4

d

)is multiplicative, we find that

(1.2) r(n) = 4

r∏

i=1

(1 + vi)

s∏

j=1

1 + (−1)wj

2.

Reader may wish to consult [2] for background on multiplicative functions, convolution of multi-plicative functions and Legendre’s symbol. Clearly, Fermat’s Theorem is an immediate corollary of(1.2).

The main object of this manuscript is to reveal new and exciting connections between the workof Ramanujan and the theory of quadratic forms. This paper is organized as follows. We collectnecessary definitions and formulas in Section 2.

In Section 3 we use 1ψ1 summation formula to prove new generalized Lambert series identities for∞∑

n,m=−∞

qn2+5m2

and

∞∑

n,m=−∞

q2n2+2nm+3m2

.

These results enable us to derive simple formulas for the number of representations of an integern by (1, 0, 5) and (2, 2, 3). Conjectures of Fermat and Euler for (1, 0, 5) are easy corollaries of theseformulas. Our treatment of (1, 0, 6) and (2, 0, 3) in Section 4 is very similar. However, in addition to 1ψ1

summation formula we need to use two cubic identities of Ramanujan. In Section 5 we treat (1, 0, 15)and (3, 0, 5). The surprise here is that we need to employ one of the fourty identities of Ramanujanfor the Rogers-Ramanujan functions. Section 6 deals with (1, 0, 27) and (4, 2, 7). We do not confineour discussion solely to BQF. In Section 7 we boldly treat quaternary forms x2 +5(y2 + z2 +w2) and5x2 + y2 + z2 + w2. We conclude with a brief description of the prospects for future work.

2. Definitions and Useful Formulas

Throughout the manuscript we assume that q is a complex number with |q| < 1. We adopt thestandard notation

(a; q)n := (1− a)(1 − aq) . . . (1 − aqn−1),

(a; q)∞ :=

∞∏

n=0

(1− aqn),

E(q) := (q; q)∞.

Next, we recall Ramanujan’s definition for a general theta function. Let

(2.1) f(a, b) :=

∞∑

n=−∞

an(n+1)/2bn(n−1)/2, |ab| < 1.

The function f(a, b) satisfies the well-known Jacobi triple product identity [6, p. 35, Entry 19]

(2.2) f(a, b) = (−a; ab)∞(−b; ab)∞(ab; ab)∞.

Two important special cases of (2.1) are

(2.3) ϕ(q) := f(q, q) =

∞∑

n=−∞

qn2

= (−q; q2)2∞(q2; q2)∞ =E5(q2)

E2(q4)E2(q),

and

(2.4) ψ(q) := f(q, q3) =∞∑

n=−∞

q2n2−n = (−q; q4)∞(−q3; q4)∞(q4; q4)∞ =

E2(q2)

E(q).

The product representations in (2.3)–(2.4) are special cases of (2.2). We shall use the famous quintupleproduct identity, which, in Ramanujan’s notation, takes the form [6, p. 80, Entry 28(iv)]

(2.5) E(q)f(−a2,−a−2q)

f(−a,−a−1q)= f(−a3q,−a−3q2) + af(−a−3q,−a3q2),

where a is any complex number.Function f(a, b) also satisfies a useful addition formula. For each nonnegative integer n, let

Un := an(n+1)/2bn(n−1)/2 and Vn := an(n−1)/2bn(n+1)/2.

Then [6, p. 48, Entry 31]

(2.6) f(U1, V1) =

n−1∑

r=0

Urf

(Un+r

Ur,Vn−r

Ur

).

From (2.6) with n = 2, we obtain

(2.7) f(a, b) = f(a3b, ab3) + af(b

a,a

b(ab)4).

A special case of (2.7) which we frequently use is

(2.8) ϕ(q) = ϕ(q4) + 2qψ(q8).

With a = b = q and n = 3, we also find that

(2.9) ϕ(q) = ϕ(q9) + 2qf(q3, q15).

Our proofs employ a well-known special case of Ramanujan’s 1ψ1 summation formula:If |q| < |a| < 1, then [6, p. 32, Entry 17]

(2.10) E3(q)f(−ab,−q/ab)

f(−a,−q/a)f(−b,−q/b)=

∞∑

n=−∞

an

1− bqn.

4 ALEXANDER BERKOVICH AND HAMZA YESILYURT

We frequently use the elementary result [6, p. 45, Entry 29]. If ab = cd, then

f(a, b)f(c, d) = f(ac, bd)f(ad, bc) + af(b

c,c

babcd)f(

b

d,d

babcd).(2.11)

Next, we recall that for an odd prime p, Legendre’s Symbol(np

)or (n | p) is defined by

(np

)=

{1 if n is a quadratic residue modulo p,−1 if n is a quadratic nonresidue modulo p.

Kronecker’s Symbol( nm

)is defined as follows

(nk

)=

1 if k = 1,0 if k is a prime dividing n,Legendre’s symbol if k is an odd prime.

(n2

)=

0 if n is even,1 if n is odd, n ≡ ±1 (mod8),−1 if n is odd, n ≡ ±3 (mod8).

In general,( nm

)=

s∏

i=1

( npi

)if m =

s∏

i=1

pi is a prime factorization of m.

It is easy to show that( abc

)=(ab

)(ac

)and

(abc

)=(ac

)(bc

). Hence,

( nm

)is a completely multi-

plicative function of n and also of m.

3. Lambert Series Identities for∑∞

n,m=−∞ qn2+5m2

Theorem 3.1.

ϕ(q)ϕ(q5) = 2{ ∞∑

n=−∞

qn

1 + q10n−

∞∑

n=−∞

q5n+2

1 + q10n+4

}(3.1)

= 2{ ∞∑

n=−∞

q3n

1 + q10n+

∞∑

n=−∞

q5n+1

1 + q10n+2

}(3.2)

= 1 +

∞∑

n=1

(−20

n

) qn

1− qn+

∞∑

n=1

(n5

) qn

1 + q2n.(3.3)

Furthermore,

(3.4) 1 +

∞∑

n=1

(−20

n

) qn

1− qn=E(q2)E(q4)E(q5)E(q10)

E(q)E(q20)

and

(3.5)

∞∑

n=1

(n5

) qn

1 + q2n= q

E(q)E(q2)E(q10)E(q20)

E(q4)E(q5).

Proof. Employing (2.10) with q, a and b replaced by q10, q and −1, respectively, we find that

(3.6)

∞∑

n=−∞

qn

1 + q10n= E3(q10)

f(q, q9)

f(−q,−q9)f(1, q10).

From (2.10), we similarly find that

(3.7)

∞∑

n=−∞

q5n+2

1 + q10n+4= q2E3(q10)

f(q, q9)

f(−q5,−q5)f(q4, q6).

From (2.11), with q replaced by q5 and with a = b = q2, we obtain

(3.8) f(−q2, q3)f(−q2, q3) = f(−q5,−q5)f(q4, q6)− q2f(1, q10)f(−q,−q9).

By (3.6), (3.7) and (3.8), we conclude that

∞∑

n=−∞

qn

1 + q10n−

∞∑

n=−∞

q5n+2

1 + q10n+4

= E3(q10)f(q, q9)

f(−q,−q9)f(1, q10)− q2E3(q10)

f(q, q9)

f(−q5,−q5)f(q4, q6)

=E3(q10)f(q, q9)

f(−q,−q9)f(1, q10)f(−q5,−q5)f(q4, q6)

{f(−q5,−q5)f(q4, q6)− q2f(1, q10)f(−q,−q9)

}

=E3(q10)f(q, q9)

f(−q,−q9)f(1, q10)f(−q5,−q5)f(q4, q6)f(−q2, q3)f(−q2, q3)

=1

2ϕ(q)ϕ(q5),

after several applications of (2.2). Similarly, we find that

∞∑

n=−∞

q3n

1 + q10n+

∞∑

n=−∞

q5n+1

1 + q10n+2

= E3(q10)f(q3, q7)

f(−q3,−q7)f(1, q10)+ qE3(q10)

f(q3, q7)

f(−q5,−q5)f(q2, q8)

=E3(q10)f(q3, q7)

f(−q3,−q7)f(1, q10)f(−q5,−q5)f(q2, q8)

{f(−q5,−q5)f(q2, q8) + qf(1, q10)f(−q3,−q7)

}

=E3(q10)f(q3, q7)

f(−q3,−q7)f(1, q10)f(−q5,−q5)f(q2, q8)f(−q, q4)f(−q, q4)

=1

2ϕ(q)ϕ(q5).

Before we move on we would like to make the following

Remark 3.2. The following generalized Lambert series identity for ϕ(q)ϕ(q5) is given in [8, cor. 6.5]

ϕ(−q)ϕ(−q5) = 2

∞∑

k=−∞

qk(5k+3)/2

1 + q5k− 2q

∞∑

k=−∞

qk(5k+7)/2

1 + q5k+2.

It would be interesting to find a direct proof that

∞∑

k=−∞

(−1)k( qk

1 + q10k−

q5k+2

1 + q10k+4

)=

∞∑

k=−∞

qk(5k+3)/2

1 + q5k− q

∞∑

k=−∞

qk(5k+7)/2

1 + q5k+2.

Next, we prove (3.3).

∞∑

n=−∞

{ q5n+1

1 + q10n+2−

q5n+2

1 + q10n+4

}

=

∞∑

n=0

{ q5n+1

1 + q10n+2−

q5n+2

1 + q10n+4−

q5n+3

1 + q10n+6+

q5n+4

1 + q10n+8

}

=

∞∑

n=1

(n5

) qn

1 + q2n.(3.9)

6 ALEXANDER BERKOVICH AND HAMZA YESILYURT

Also,

∞∑

n=−∞

qn + q3n

1 + q10n

= 1 +∑

j∈{1,3,7,9}

∞∑

n=1

qjn

1 + q10n

= 1 +∑

j∈{1,3,7,9}

∞∑

n=1

∞∑

m=0

(−1)mqjnq10nm

= 1 +∑

j∈{1,3,7,9}

∞∑

n=1

∞∑

m=0

(−1)mqn(10m+j)

= 1 +∑

j∈{1,3,7,9}

∞∑

m=0

(−1)mq10m+j

1− q10m+j

= 1 +

∞∑

n=1

(−20

n

) qn

1− qn.(3.10)

Now using (3.9) and (3.10) together with (3.1) and (3.2), we see that (3.3) is proved.The equations (3.4) and (3.5) are essentially given in [17, eqs. 3.2, 3.29]. Moreover, the eta-quotients

that appear in these equations are included in a list of certain multiplicative functions determinedby Y. Martin [13]. We should emphasize that (3.1)–(3.3) are new. We observe directly that thecoefficients of the two Lambert series in (3.3) are multiplicative and that they differ at most by a sign.This enables us to compute the coefficients of ϕ(q)ϕ(q5).

Corollary 3.3. Let a(n) be the number of representations of a positive integer n by quadratic form

k2 + 5l2. If the prime factorization of n is given by

n = 2a5br∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1, 3, 7, or 9 (mod 20) and qi ≡ 11, 13, 17, or 19 (mod 20), then

(3.11) a(n) =(1 + (−1)a+t

) r∏

i=1

(1 + vi)

s∏

j=1

1 + (−1)wj

2,

where t is the number of prime factors of n, counting multiplicity, that are congruent to 3 or 7(mod 20).

Proof. Observe that

∞∑

n=1

(n5

) qn

1 + q2n=

∞∑

n=1

∞∑

m=0

(−1)m(n5

)qn(2m+1) =

∞∑

n=1

∞∑

m=1

(−4

m

)(n5

)qnm

=

∞∑

n=1

(∑

d|n

(−4

d

)(n/d5

))qn.(3.12)

Similarly,

(3.13)

∞∑

n=1

(−20

n

) qn

1− qn=

∞∑

n=1

(∑

d|n

(−20

d

))qn.

Define

(3.14) b(n) =∑

d|n

(−20

d

)and c(n) =

d|n

(−4

d

)(n/d5

).

We have by (3.3) that a(n) = b(n) + c(n). Clearly both b(n) and c(n) are multiplicative functions.Therefore, one only needs to find their values at prime powers. It is easy to check that for a prime p

(3.15) b(pα) =

1 if p = 2 or 5,1 + α if p ≡ 1, 3, 7, 9 (mod20),1 + (−1)α

2if p ≡ 11, 13, 17, 19 (mod20),

and

(3.16) c(pα) =

(−1)α if p = 2,1 if p = 5,1 + α if p ≡ 1, 9 (mod20),(−1)α(1 + α) if p ≡ 3, 7 (mod20),1 + (−1)α

2if p ≡ 11, 13, 17, 19 (mod20).

Equivalent reformulations of (3.11) can also be found in [10, p. 84, ex. 1] and [12, Thr. 7]. We shouldremark that (3.11) implies conjectures of Fermat and Euler for (1, 0, 5) stated in the introduction. Thelast two equations immediately imply (3.11).

We now determine the representations of integers by the quadratic form (2, 2, 3) and make somefurther observations.

Corollary 3.4. Let d(n) be a number of representations of a positive integer n by the quadratic form

2k2 + 2kl+ 3l2. If the prime factorization of n is given by

n = 2a5br∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1, 3, 7, or 9 (mod 20) and qi ≡ 11, 13, 17, or 19 (mod 20), then

(3.17) d(n) =(1− (−1)a+t

) r∏

i=1

(1 + vi)s∏

j=1

1 + (−1)wj

2,

where t is the number of prime factors of n, counting multiplicity, that are congruent to 3 or 7 (mod20).

Proof. Recall that

∞∑

n=0

a(n)qn :=

∞∑

k, l=−∞

qk2+5l2 .

By comparing (3.11) and (3.17), it suffices to show that d(n) = a(2n) for all n ∈ N. To that end weobserve

8 ALEXANDER BERKOVICH AND HAMZA YESILYURT

∞∑

n=0

d(n)qn =

∞∑

n,m=−∞

q2n2+2nm+3m2

=

∞∑

n,m=−∞

q2n2+2n(2m)+3(2m)2 +

∞∑

n,m=−∞

q2n2+2n(2m+1)+3(2m+1)2

=

∞∑

n,m=−∞

q2((n+m)2+5m2

)+

∞∑

n,m=−∞

q(2n+2m+1)2+5(2m+1)2

2

=∞∑

n,m=−∞

q(2n)2+5(2m)2

2 +∞∑

n,m=−∞

q(2n+1)2+5(2m+1)2

2

=

∞∑

n=0

a(2n)qn.

By (3.3), (3.11) and (3.17) we find that

∞∑

n,m=−∞

q2n2+2nm+3m2

= 1 +

∞∑

n=1

(−20

n

) qn

1− qn−

∞∑

n=1

(n5

) qn

1 + q2n.(3.18)

Also by adding identities in (3.3) and (3.18), we conclude that

∞∑

n,m=−∞

qn2+5m2

+

∞∑

n,m=−∞

q2n2+2nm+3m2

= 2 + 2

∞∑

n=1

(−20

n

) qn

1− qn.

This last equation is a special case of Dirichlet’s formula [18, p. 123, thr. 4]. Comparing (3.11) and(3.17) we see that a(n)d(n) = 0. This means that a positive integer cannot be represented by (1,0,5)and (2,2,3) at the same time.

We end this section by proving a Lambert series representation for ψ(q)ψ(q5).

Theorem 3.5.

(3.19) ψ(q)ψ(q5) =∞∑

n=−∞

q3n + q7n+1

1− q20n+5=

∞∑

n=−∞

qn + q9n+6

1− q20n+15.

Proof. By two applications of (2.10) with q replaced by q20, a = q3, q7 and b = q5, we find that

∞∑

n=−∞

q3n + q7n+1

1− q20n+5= E3(q20)

f(−q8,−q12)

f(−q3,−q17)f(−q5,−q15)+ qE3(q20)

f(−q8,−q12)

f(−q7,−q13)f(−q5,−q15)

=E3(q20)f(−q8,−q12)

f(−q5,−q15)f(−q3,−q17)f(−q7,−q13)

{f(−q7,−q13) + qf(−q3,−q17)

}

=E3(q20)f(−q8,−q12)

f(−q5,−q15)f(−q3,−q17)f(−q7,−q13)f(q,−q4),(3.20)

where in the last step we use (2.7) with a = q and b = −q4. It is now easy to verify by severalapplications of (2.2) that (3.20) is equal to ψ(q)ψ(q5). The proof of the second representation givenin (3.19) is very similar to that of the first one and so we forego its proof. �

4. Lambert Series Identities for∑∞

n,m=−∞ qn2+6m2

and∑∞

n,m=−∞ q2n2+3m2

Theorem 4.1. Let

(4.1) P (q) :=E(q2)E(q3)E(q8)E(q12)

E(q)E(q24)and Q(q) := q

E(q)E(q4)E(q6)E(q24)

E(q3)E(q8).

Then,

P (q) =

∞∑

n=−∞

qn + q5n

1 + q12n= 1 +

∞∑

n=1

(−6

n

) qn

1− qn,(4.2)

Q(q) =

∞∑

n=−∞

q3n+1 − q9n+3

1 + q12n+4=

∞∑

n=1

(n3

)qn(1 − q2n)

1 + q4n.(4.3)

Moreover,

ϕ(q)ϕ(q6) = P (q) +Q(q)(4.4)

=

∞∑

n=−∞

qn + q5n

1 + q12n+

∞∑

n=−∞

q3n+1 − q9n+3

1 + q12n+4(4.5)

= 2{ ∞∑

n=−∞

qn

1 + q12n−

∞∑

n=−∞

q9n+3

1 + q12n+4

}(4.6)

= 2{ ∞∑

n=−∞

q5n

1 + q12n+

∞∑

n=−∞

q3n+1

1 + q12n+4

}(4.7)

= 1 +

∞∑

n=1

(−6

n

) qn

1− qn+

∞∑

n=1

(n3

)qn(1 − q2n)

1 + q4n(4.8)

and

ϕ(q2)ϕ(q3) = P (q)−Q(q)(4.9)

=∞∑

n=−∞

qn + q5n

1 + q12n−

∞∑

n=−∞

q3n+1 − q9n+3

1 + q12n+4(4.10)

= 2{ ∞∑

n=−∞

qn

1 + q12n−

∞∑

n=−∞

q3n+1

1 + q12n+4

}(4.11)

= 2{ ∞∑

n=−∞

q5n

1 + q12n+

∞∑

n=−∞

q9n+3

1 + q12n+4

}(4.12)

= 1 +∞∑

n=1

(−6

n

) qn

1− qn−

∞∑

n=1

(n3

)qn(1 − q2n)

1 + q4n.(4.13)

Proof. By employing (2.11) with a = q, b = q11, c = −q5 and d = −q7, we find that

f(q, q11)f(−q5,−q7) = f(−q6,−q18)f(−q8,−q16) + qf(−q4,−q20)f(−q6,−q18)

= ψ(−q6)E(q8) + qψ(−q6)f(−q4,−q20).(4.14)

10 ALEXANDER BERKOVICH AND HAMZA YESILYURT

By two applications of (2.10) with q replaced by q12, a = q, q5, b = −1, and by (4.14), we find that∞∑

n=−∞

qn + q5n

1 + q12n= E3(q12)

f(q, q11)

f(1, q12)f(−q,−q11)+ E3(q12)

f(q5, q7)

f(1, q12)f(−q5,−q7)

=E3(q12)

f(1, q12)f(−q,−q11)f(−q5,−q7)

{f(q, q11)f(−q5,−q7) + f(−q,−q11)f(q5, q7)

}

= 2E3(q12)

f(1, q12)f(−q,−q11)f(−q5,−q7)ψ(−q6)E(q8)

=E(q2)E(q3)E(q8)E(q12)

E(q)E(q24),(4.15)

after several applications of (2.2).By (2.7), we observe that

(4.16) E(q) = f(−q,−q2) = f(q5, q7)− qf(q, q11).

Arguing as above and using (4.16), we conclude that∞∑

n=−∞

q3n+1 − q9n+3

1 + q12n+4= qE3(q12)

f(q5, q7)

f(−q3,−q9)f(q4, q8)− q3E3(q12)

f(q−1, q13)

f(−q3,−q9)f(q4, q8)

= qE3(q12)f(q5, q7)

f(−q3,−q9)f(q4, q8)− q2E3(q12)

f(q, q11)

f(−q3,−q9)f(q4, q8)

= qE3(q12)

ψ(−q3)f(q4, q8)

{f(q5, q7)− qf(q, q11)

}

= qE3(q12)E(q)

ψ(−q3)f(q4, q8)= q

E(q)E(q4)E(q6)E(q24)

E(q3)E(q8),(4.17)

after several applications of (2.2).The proofs of second part of (4.2) and that of (4.3) are similar to those of (3.9), (3.10) and so we

omit their proofs.We prove (4.4) and (4.9) simultaneously by proving

2P = ϕ(q)ϕ(q6) + ϕ(q2)ϕ(q3)(4.18)

and

2Q = ϕ(q)ϕ(q6)− ϕ(q2)ϕ(q3).(4.19)

First we prove (4.19). We will need two identities of Ramanujan [6, p. 232], namely

(4.20) 2ψ3(q)

ψ(q3)=ϕ3(q)

ϕ(q3)+ϕ3(−q2)

ϕ(−q6)

and

(4.21) 4qψ(q2)ψ(q6) = ϕ(q)ϕ(q3)− ϕ(−q)ϕ(−q3).

From (4.21) with (2.8), we find that

4qψ(q2)ψ(q6) = ϕ(q)ϕ(q3)− ϕ(−q)ϕ(−q3)

=(ϕ(q4) + qψ(q8)

)(ϕ(q12) + q3ψ(q24)

)−(ϕ(q4)− qψ(q8)

)(ϕ(q12)− q3ψ(q24)

)

= 4q{ψ(q8)ϕ(q12) + q2ϕ(q4)ψ(q24)

}.(4.22)

Upon replacing q2 by q in (4.22), we conclude that

(4.23) ψ(q)ψ(q3) = ψ(q4)ϕ(q6) + qϕ(q2)ψ(q12).

Similarly,

ϕ(q)ϕ(−q3)− ϕ(−q)ϕ(q3)

=(ϕ(q4) + qψ(q8)

)(ϕ(q12)− q3ψ(q24)

)−(ϕ(q4)− qψ(q8)

)(ϕ(q12) + q3ψ(q24)

)

= 4q{ψ(q8)ϕ(q12)− q2ϕ(q4)ψ(q24)

}

= 4qψ(−q2)ψ(−q6),(4.24)

where in the last step we used (4.23) with q replaced by −q2. We are now ready to prove (4.19).Recall that Q(q) is defined by (4.1). By several applications of (2.2), we see that (4.19) is equivalentto

(4.25) 2qψ(−q)ψ(−q2)ψ(−q3)ψ(−q6)

ψ(q4)ϕ(−q3)= ϕ(q)ϕ(q6)− ϕ(q2)ϕ(q3),

or

2qψ(−q)ψ(−q2)ψ(−q3)ψ(−q6) = ϕ(q)ϕ(q6)ψ(q4)ϕ(−q3)− ϕ(q2)ϕ(q3)ψ(q4)ϕ(−q3)

= ϕ(q)ϕ(−q3)ψ(q4)ϕ(q6)− ψ2(q2)ϕ2(−q6),(4.26)

where we used the trivial identities

(4.27) ϕ(q)ϕ(−q) = ϕ2(−q2) and ψ2(q) = ψ(q2)φ(q).

When we employ (4.24) on the far left hand side of (4.26) and (4.23) on the right hand side of(4.26), we find that

(4.28)ψ(−q)ψ(−q3)

(ϕ(q)ϕ(−q3)−ϕ(−q)ϕ(q3)

)=(ψ(q)ψ(q3)+ψ(−q)ψ(−q3)

)ϕ(q)ϕ(−q3)−2ψ2(q2)ϕ2(−q6).

Upon cancellation, we see that

(4.29) −ψ(−q)ψ(−q3)ϕ(−q)ϕ(q3) = ψ(q)ψ(q3)ϕ(q)ϕ(−q3)− 2ψ2(q2)ϕ2(−q6).

Next we multiply both sides of (4.29) withϕ2(q)

ψ(q)ψ(q3)ϕ2(−q6)and obtain after several applications

of (2.2) that

(4.30) −ϕ3(−q2)

ϕ(−q6)=ϕ3(q3)

ϕ(q3)− 2

ψ3(q)

ψ(q3),

which is (4.20). Hence the proof of (4.19) is complete.The proof of (4.18) is very similar to that of (4.19). Recall that R(q) is defined by (4.1). By several

applications of (2.2), we see that (4.19) is equivalent to

(4.31) 2ψ(−q)ψ(−q2)ψ(−q3)ψ(−q6)

ψ(q12)ϕ(−q)= ϕ(q)ϕ(q6) + ϕ(q2)ϕ(q3),

or

ψ(−q)ψ(−q2)ψ(−q3)ψ(−q6) = ϕ(q)ϕ(q6)ψ(q12)ϕ(−q) + ϕ(q2)ϕ(q3)ψ(q12)ϕ(−q)

= ϕ2(−q2)ψ2(q6) + ϕ(−q)ϕ(q3)ϕ(q2)ψ(q12).(4.32)

If we employ (4.24) on the far left hand side of (4.32), and (4.23) on the right hand side of (4.32)and multiply both sides by 2q, we find that

(4.33)ψ(−q)ψ(−q3)

(ϕ(q)ϕ(−q3)−ϕ(−q)ϕ(q3)

)= 2qψ2(q6)ϕ2(−q2)+ϕ(−q)ϕ(q3)

(ψ(q)ψ(q3)−ψ(−q)ψ(−q3)

).

Upon cancellation, we find that

(4.34) ψ(−q)ψ(−q3)ϕ(q)ϕ(−q3) = ψ(q)ψ(q3)ϕ(−q)ϕ(q3) + 2qψ2(q6)ϕ2(−q2).

12 ALEXANDER BERKOVICH AND HAMZA YESILYURT

It is easy to see that (4.34) and (4.29) are “reciprocals” of each other. For related definitions andmodular equations corresponding to (4.18) and (4.19) see [6, p. 230, Entry 5 (i)]. Hence, the proof of(4.18) is complete.

As an immediate corollary of (4.18) and (4.19), we note the following two interesting theta functionidentities

(4.35)ϕ(q)ϕ(q6)− ϕ(q2)ϕ(q3)

ϕ(q)ϕ(q6) + ϕ(q2)ϕ(q3)= q

ϕ(−q)ψ(q12)

ϕ(−q3)ψ(q4)

and

ϕ2(q)ϕ2(q6)− ϕ2(q2)ϕ2(q3) = 4qE(q2)E(q4)E(q6)E(q12) = 4qψ(q)ψ(−q)ψ(−q3)ψ(−q6).

The identities (4.2) and (4.3) together with (4.4) and (4.9) clearly imply (3.1) and (3.3). To provethe remaining identities (4.6), (4.7), (4.11) and (4.12), one only needs to prove that

(4.36)∞∑

n=−∞

qn − q5n

1 + q12n=

∞∑

n=−∞

q3n+1 + q9n+3

1 + q12n+4.

Arguing as in (4.15) and (4.17), one can easily show that

(4.37)∞∑

n=−∞

qn − q5n

1 + q12n=

∞∑

n=−∞

q3n+1 + q9n+3

1 + q12n+4=E(q2)E(q3)E(q4)E(q24)

E(q)E(q8).

Hence, the proof of the Theorem 4.1 is complete. �

Corollary 4.2. Let a(n) and b(n) be the number of representations of a positive integer n by quadratic

form k2 + 6l2 and 2k2 + 3l2, respectively. If the prime factorization of n is given by

n = 2a3br∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1, 5, 7, or 11 (mod 24) and qi ≡ 13, 17, 19, or 23 (mod 24), then

(4.38) a(n) =(1 + (−1)a+b+t

) r∏

i=1

(1 + vi)

s∏

j=1

1 + (−1)wj

2

and

(4.39) b(n) =(1− (−1)a+b+t

) r∏

i=1

(1 + vi)s∏

j=1

1 + (−1)wj

2,

where t is a number of prime factors of n, counting multiplicity, that are congruent to 5 or 11 (mod 24).

Proof. Observe that∞∑

n=1

(n3

)qn(1− q2n)

1 + q4n=

∞∑

n=1

∞∑

m=0

(−1)m(n3

)(qn(4m+1) − qn(4m+3))

=

∞∑

n=1

∞∑

m=1

(−1)m(n3

)(qn(4m−1) − qn(4m−3))

=

∞∑

n=1

∞∑

m=1

(m2

)(n3

)qnm

=

∞∑

n=1

(∑

d|n

(d2

)(n/d3

))qn.

Similarly,∞∑

n=1

(−6

n

) qn

1− qn=

∞∑

n=1

(∑

d|n

(−6

d

))qn.

Define

c(n) =∑

d|n

(−24

d

)and d(n) =

d|n

(d2

)(n/d3

).

Equations (4.8) and (4.13) imply that a(n) = c(n) + d(n) and b(n) = c(n)− d(n). Clearly both c(n)and d(n) are multiplicative functions. Therefore, one only needs to find their values at prime powers.It is easy to check that for a prime p

(4.40) c(pα) =

1 if p = 2 or 3,1 + α if p ≡ 1, 5, 7, 11 (mod24),1 + (−1)α

2if p ≡ 13, 17, 19, 23 (mod24)

,

and

(4.41) d(pα) =

(−1)α if p = 2 or 3,1 + α if p ≡ 1, 7 (mod24),(−1)α(1 + α) if p ≡ 5, 11 (mod24),1 + (−1)α

2if p ≡ 13, 17, 19, 23 (mod24).

From these two equations (4.38) and (4.39) are immediate. Equivalent reformulations of (4.38) and(4.39) can also be found in [10, p. 84, ex. 2] and [12, Thr. 7]. �

5. Lambert Series Identities for∑∞

n,m=−∞ qn2+15m2

and∑∞

n,m=−∞ q3n2+5m2

Theorem 5.1. Let

(5.1) P (q) :=E(q)E(q6)E(q10)E(q15)

E(q2)E(q30)and Q(q) := q

E(q2)E(q3)E(q5)E(q30)

E(q6)E(q10).

Then,

P (q) = 1−

∞∑

n=1

(−15

n

) qn

1 + qn,(5.2)

Q(q) =

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n.(5.3)

Moreover,

ϕ(−q)ϕ(−q15) = P (q)−Q(q)(5.4)

= 1−

∞∑

n=1

(−15

n

) qn

1 + qn−

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n(5.5)

and

ϕ(−q3)ϕ(−q5) = P (q) +Q(q)(5.6)

= 1−

∞∑

n=1

(−15

n

) qn

1 + qn+

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n.(5.7)

14 ALEXANDER BERKOVICH AND HAMZA YESILYURT

Proof. The identities (5.2), (5.4), and (5.6) were observed by Ramanujan [6, p. 379, Entry 10 (vi)],[21, eq. 50] and [6, p. 377, Entry 9 (v), (vi)]. We prove (5.3).It is easy to observe that

(5.8)

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n=

∞∑

n=−∞

q5n+1 + q10n+2

1 + q15n+3−

∞∑

n=−∞

q5n+1 + q10n+2

1 + q15n+3.

Next by four applications of (2.10) on the right hand side of (5.8), we have

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n= E3(q15)

{q

f(q7, q8)

E(q5)f(q3, q12)+ q2

f(q2, q13)

E(q5)f(q3, q12)

}

− E3(q15){q2

f(q4, q11)

E(q5)f(q6, q9)+ q3

f(q, q14)

E(q5)f(q6, q9)

}

= qE3(q15)

E(q5)f(q3, q12)f(q6, q9)

{(f(q7, q8) + qf(q2, q13)

)f(q6, q9)

− q(f(q4, q11) + qf(q, q14)

)f(q3, q12)

}.(5.9)

Now we employ (2.11) for each term of (5.9) inside the parenthesis, the identity in (5.9) now becomes

qE3(q15)

E(q5)f(q3, q12)f(q6, q9)

{(f(q14, q16)− q2f(q4, q26)

)(f(q13, q17)− qf(q7, q23)

)

+ q(f(q8, q22)− q2f(q2, q28)

)(f(q11, q19)− q3f(q, q29)

).(5.10)

Recall that the Rogers-Ramanujan functions are defined by

(5.11) G(q) :=

∞∑

n=0

qn2

(q; q)nand H(q) :=

∞∑

n=0

qn(n+1)

(q; q)n.

These functions satisfy the famous Rogers–Ramanujan identities [16, pp. 214–215]

(5.12) G(q) =1

(q; q5)∞(q4; q5)∞and H(q) =

1

(q2; q5)∞(q3; q5)∞.

Our proof makes use of one of Ramanujan’s forty identities for the Rogers-Ramanujan functions,namely [20]

(5.13) G(q)G(q4)− qH(q)H(q4) =ϕ(q5)

E(q2).

Next, we employ the quintuple product identity (2.5), with q replaced by q10 and a = −q to findthat

(5.14) f(−q13,−q17) + qf(−q7,−q23) = E(q10)f(−q2,−q8)

f(−q,−q9)= E(q2)G(q).

Similarly, from (2.5) we find

E(q2)H(q) = f(−q11,−q19) + q3f(−q,−q29),(5.15)

E(q)G(q2) = f(q7, q8)− qf(q2, q13),(5.16)

E(q)H(q2) = f(q4, q11)− qf(q, q14).(5.17)

Next, making use of (5.8)-(5.10), (5.14)–(5.17), (5.13), and (5.1), we conclude that

∞∑

n=1

( 5

n

)qn(1 + q2n)

1 + q3n

= qE3(q15)

E(q5)f(q3, q12)f(q6, q9)E2(q2)

{G(q4)G(−q) + qH(q4)H(−q)

}

= qE3(q15)

E(q5)f(q3, q12)f(q6, q9)E2(q2)

ϕ(−q5)

E(q2)

= qE(q2)E(q3)E(q5)E(q30)

E(q6)E(q10)

= Q(q).

Adding together (5.4) and (5.6) and replacing q by −q, we find that

ϕ(q)ϕ(q15) + ϕ(q3)ϕ(q5) = 2− 2

∞∑

n=1

(−15

n

) (−q)n

1 + (−q)n.

It is instructive to compare this formula with an equivalent formula (50) in [21] which states that

ϕ(q)ϕ(q15) + ϕ(q3)ϕ(q5) = 2 +

∞∑

n=1

a(n)qn

1− qn,

where

a(n) = 2(−60

n

)−2δ(2|n)

(−60

n/2

)+2δ(4|n)

(−15

n/4

),

with

δ(a|b) =

{1 if a|b,0 otherwise .

Corollary 5.2. Let a(n) and b(n) be the number of representations of a positive integer n by quadratic

form k2 + 15l2 and 3k2 + 5l2, respectively. If the prime factorization of n is given by

n = 2a3b5cr∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1, 2, 4, or 8 (mod 15) and qi ≡ 7, 11, 13, or 14 (mod 15), then

(5.18) a(n) = |a− 1|(1 + (−1)a+b+c+t

) r∏

i=1

(1 + vi)s∏

j=1

1 + (−1)wj

2

and

(5.19) b(n) = |a− 1|(1− (−1)a+b+c+t

) r∏

i=1

(1 + vi)s∏

j=1

1 + (−1)wj

2,

where t is a number of prime factors of n, counting multiplicity, that are congruent to 2 or 8 (mod 15).

16 ALEXANDER BERKOVICH AND HAMZA YESILYURT

Proof. Observe that

Q(q) =

∞∑

n=1

( 5

n

)qn(1 + qn)

1 + q3n=

∞∑

n=1

∞∑

m=0

(−1)m( 5

n

)(qn(3m+1) + qn(3m+2))

= −

∞∑

n=1

∞∑

m=1

(−1)m( 5

n

)(qn(3m−1) + qn(3m−2))

= −

∞∑

n=1

∞∑

m=1

(−1)m(−3

m

)( 5

n

)qnm

= −

∞∑

n=1

(∑

d|n

(−1)d(−3

d

)( 5

n/d

))qn.

Therefore,

(5.20) Q(−q) = −

∞∑

n=1

(∑

d|n

(−1)n+d(−3

d

)( 5

n/d

))qn.

Similarly,

P (−q) = 1 +∞∑

n=1

(∑

d|n

(−1)n+d(−15

n/d

))qn.

Now we define

c(n) =∑

d|n

(−1)n+d(−15

n/d

)and d(n) =

d|n

(−1)n+d(−3

d

)( 5

n/d

).

By (5.4) and (5.6) we have a(n) = c(n) + d(n) and b(n) = c(n)− d(n), for n > 0. Actually, c(n) andd(n) are multiplicative functions. This fact is an easy corollary of the following

Lemma 5.3. If α(n) and β(n) are multiplicative then so is

F (n) :=∑

d|n

α(n/d)β(d)(−1)n+d.

Proof. The proof is a straightforward case analysis. �

Since c(n) and d(n) are multiplicative functions, one only needs to find their values at prime powers.It is easy to check that for a prime p

(5.21) c(pα) =

a− 1 if p = 2,1 if p = 3 or 5,1 + α if p ≡ 1, 2, 4, 8 (mod15),1 + (−1)α

2if p ≡ 7, 11, 13, 14 (mod15),

and

(5.22) d(pα) =

(−1)α(a− 1) if p = 2,(−1)α if p = 3 or 5,1 + α if p ≡ 1, 4 (mod15),(−1)α(1 + α) if p ≡ 1, 4 (mod15),1 + (−1)α

2if p ≡ 7, 11, 13, 14 (mod15).

From these two equations (5.18) and (5.19) are immediate. Equivalent reformulations of (5.18) and(5.19) can also be found in [12]. �

6. Representations by the the quadratic form k2 + 27l2

In this section, we give a formula for the number of representations of a positive integer by thequadratic form k2+27l2. First we state the following theorem [22, thr. (3.1), cor. (3.3)] which we willuse in our proof.

Theorem 6.1. Let m be an integer and α, β, p and λ be positive integers such that

(6.1) αm2 + β = pλ.

Let δ, ǫ be integers. Further let l and t be real and x and y be nonzero complex numbers.

We define

(6.2) fk(a, b) =

{f(a, b) if k ≡ 0 (mod 2),f(−a,−b) if k ≡ 1 (mod 2).

With the parameters defined this way, we set

R(ǫ, δ, l, t, α, β,m, p, λ, x, y)

:=

p−1∑

k=0n=2k+t

(−1)ǫkykq{λn2+pαl2+2αnml}/4fδ(xq

(1+l)pα+αnm, x−1q(1−l)pα−αnm)

× fǫp+mδ(x−mypqpβ+βn, xmy−pqpβ−βn).(6.3)

Then,

R(ǫ, δ, l, t, α, β,m, p, λ, x, y) =

∞∑

u, v=−∞

(−1)δv+ǫuxvyuqλ(2u+t)2+2αm(2u+t)(2v+l)+pα(2v+l)2

4 .(6.4)

Moreover, let α1, β1, m1, p1 be another set of parameters such that

α1m21 + β1 = p1λ, αβ = α1β1 and λ|(αm− α1m1). Set a :=

αm− α1m1

λ. Then,

(6.5) R(ǫ, δ, l, t, α, β,m, p, λ, x, y) = R(ǫ, δ + aǫ, l, t+ al, α1, β1,m1, p1, λ, xy−a, y).

We are now ready to state the results of this section

Theorem 6.2.

(6.6) ϕ(q)ϕ(q27) =ϕ(q)ϕ(q3)− ϕ(q3)ϕ(q9)

3+ ϕ(q9)ϕ(q27) +

4

3qE(q6)E(q18).

Let a(n), b(n) be the number of representations of a positive integer n by quadratic forms (1, 0, 27)and (4, 2, 7), respectively.If n 6≡ 1 (mod 6), then(6.7)

a(n) = b(n) =

(3− 2δα,0

)(1 + (−1)α)

∏ri=1(1 + vi)

∏sj=1

1 + (−1)wj

2if β ≥ 2,

(1 + (−1)α)∏r

i=1(1 + vi)∏s

j=1

1 + (−1)wj

2if β = 0 and α > 0,

0 otherwise,

where n has a prime factorization

n = 2α3βr∏

i=1

pvii

s∏

j=1

qwj

j ,

with pi ≡ 1 (mod 3) and 2 6= qi ≡ 2 (mod 3) and

(6.8) δj,0 :=

{1 if j = 0,0 otherwise.

If n ≡ 1 (mod6), then

18 ALEXANDER BERKOVICH AND HAMZA YESILYURT

a(n) =2

3

r∏

i=1

(1 + vi)( s∏

i=1

(1 + ui) + 2

s∏

i=1

((1 + ui) | 3

)) t∏

i=1

1 + (−1)wi

2,(6.9)

b(n) =2

3

r∏

i=1

(1 + vi)( s∏

i=1

(1 + ui)−

s∏

i=1

((1 + ui) | 3)

)) t∏

i=1

1 + (−1)wi

2,(6.10)

where n has a prime factorization

(6.11)

r∏

i=1

pvii

s∏

i=1

qui

i

t∏

i=1

Qwi

j ,

with pi ≡ 1 (mod 3), 2pi−13 ≡ 1 (mod pi), qi ≡ 1 (mod 3), 2

qi−13 6≡ 1 (mod qi) and 2 6= Qi ≡ 2 (mod 3).

Proof. From (6.5) together with (6.3), we find that

R(0, 0, 0, 0, 1, 27, 1, 4, 7, 1, 1) = f(q4, q4)f(q108, q108) + 2q7f(q2, q6)f(q54, q162) + q28f(1, q8)f(1, q216)

(6.12)

= (ϕ(q)ϕ(q27) + ϕ(−q)ϕ(−q27))/2 + 2q7ψ(q2)ψ(q54),(6.13)

where in the last step we used (2.8).Similarly,

R(1, 0, 1, 0, 1, 27, 1, 4, 7, 1, 1) = qf(1, q8)f(q108, q108)− 2q7f(q2, q6)f(q54, q162) + q27f(q4, q4)f(1, q216)

(6.14)

= (ϕ(q)ϕ(q27)− ϕ(−q)ϕ(−q27))/2− 2q7ψ(q2)ψ(q54).(6.15)

Moreover,

R(1, 0, 1, 0, 1, 27, 1, 4, 7, 1, 1) = R(1, 1, 1, 1, 3, 9,−2, 3, 7, 1, 1)(6.16)

= 2qf(−q6,−q12)f(−q18,−q36)(6.17)

= 2qE(q6)E(q18).(6.18)

Using (2.9) we get

R(0, 0, 0, 0, 1, 27, 1, 4, 7, 1, 1) = R(0, 0, 0, 0, 3, 9,−2, 3, 7, 1, 1)(6.19)

= f(q9, q9)f(q27, q27) + 2q4f(q3, q15)f(q9, q45)(6.20)

= ϕ(q9)ϕ(q27) + (ϕ(q) − ϕ(q9))(ϕ(q3)− ϕ(q27))/2(6.21)

=(3ϕ(q9)ϕ(q27) + ϕ(q)ϕ(q3)− ϕ(q)ϕ(q27)− ϕ(q3)ϕ(q9)

)/2.(6.22)

Adding (6.13) and (6.15) and using the alternate representations (6.18) and (6.22), we find that

(6.23) ϕ(q)ϕ(q27) =(3ϕ(q9)ϕ(q27) + ϕ(q)ϕ(q3)− ϕ(q)ϕ(q27)− ϕ(q3)ϕ(q9)

)/2 + 2qE(q6)E(q18),

which is (6.6). Recall that

(6.24) 1 +

∞∑

n=1

a(n)qn := ϕ(q)ϕ(q27) and 1 +

∞∑

n=1

b(n)qn :=

∞∑

n,m=−∞

q4n2+2nm+7m2

.

We also define c(n) and d(n) by

(6.25) 1 +∞∑

n=1

c(n)qn = ϕ(q)ϕ(q3) and∞∑

n=1

d(n)qn = qE(q6)E(q18).

Using the following Lambert series expansion for ϕ(q)ϕ(q3) (see for example [7, p. 75, eq. (3.7.8)])

(6.26) ϕ(q)ϕ(q3) = 1 + 2∞∑

n=1

(n3

) qn

1− qn+ 4

∞∑

n=1

(n3

) q4n

1− q4n,

it is easy to show that

(6.27) c(n) =(3− 2δα,0

)(1 + (−1)α)

r∏

i=1

(1 + vi)s∏

j=1

1 + (−1)wj

2,

where n has a prime factorization

n = 2α3βr∏

i=1

pvii

s∏

j=1

qwj

j ,

where pi ≡ 1 (mod3) and 2 6= qi ≡ 2 (mod3).From (6.6), we have

(6.28) a(n) =c(n)− c(n/3)

3+ c(n/9) +

4

3d(n),

where we assume c(n/l) = 0 if l 6 |n. If n 6≡ 1 (mod6), then d(n) ≡ 0 and a(n) = c(n/9) if 3|n, whilea(n) = c(n)/3 if 3 6 |n. This proves the claim in (6.7) for a(n).

Now assume n ≡ 1 (mod6). If p is a prime and p ≡ 1 (mod 3), then by (6.27) we see that c(p) = 4.If p is represented by the form (1, 0, 27), then a(p) = 4. This is because 4 ≤ a(p) ≤ c(p) = 4. Using(6.28) we see that d(p) = 2. If p is not represented by (1, 0, 27) then, by (6.28), d(p) = −1. Gaussproved that if p is a prime and p ≡ 1 (mod 3), then p is represented by (1, 0, 27) iff 2 is a cubic residuemodulo p or, equivalently, iff 2(p−1)/3 ≡ 1 (mod p). Therefore,

(6.29) d(p) =

2 if p ≡ 1 (mod3) and 2(p−1)/3 ≡ 1 (mod p)−1 if p ≡ 1 (mod3) and 2(p−1)/3 6≡ 1 (mod p)0 if p 6≡ 1 (mod3).

In [13], Y. Martin proved that qE(q6)E(q18) is a multiplicative cusp form in S1

(Γ0(108),

(−108

)).

That is d(n) is multiplicative and for any prime p and s > 0

(6.30) d(qs+2) = d(p)d(ps+1)−(−108

p

)d(ps),

where d(1) = 1. Using this recursion formula together with (6.29), we find that

(6.31) d(pα) =

α+ 1 if p ≡ 1 (mod3) and 2(p−1)/3 ≡ 1 (mod p),((α+ 1)|3

)if p ≡ 1 (mod3) and 2(p−1)/3 6≡ 1 (mod p),

(1 + (−1)α)/2 if 2 6= p ≡ 2 (mod 3),0 if p = 2 or 3.

Therefore,

(6.32) d(n) = δα,0δβ,0

r∏

i=1

(1 + vi)

s∏

i=1

((1 + ui) | 3)

) t∏

i=1

1 + (−1)wi

2,

where n has a prime factorization

(6.33) 2α3βr∏

i=1

pvii

s∏

i=1

qui

i

t∏

i=1

Qwi

j ,

where pi ≡ 1 (mod 3), 2pi−13 ≡ 1 (mod pi), qi ≡ 1 (mod3), 2

qi−13 6≡ 1 (mod qi) and 2 6= Qi ≡ 2 (mod 3).

By (6.28), if n ≡ 1 (mod6), then a(n) =c(n) + 4d(n)

3. Using (6.27) and (6.32) we arrive at the

20 ALEXANDER BERKOVICH AND HAMZA YESILYURT

statement for a(n) given in (6.9). Next, by (6.13), we have

1 +

∞∑

n=1

b(n)qn : =

∞∑

n,m=−∞

q4n2+2nm+7m2

(6.34)

= R(0, 0, 0, 0, 1, 27, 1, 4, 7, 1, 1)(6.35)

= (ϕ(q)ϕ(q27) + ϕ(−q)ϕ(−q27))/2 + 2q7ψ(q2)ψ(q54).(6.36)

Therefore,

ϕ(q)ϕ(q27)−

∞∑

n,m=−∞

q4n2+2nm+7m2

(6.37)

= (ϕ(q)ϕ(q27)− ϕ(−q)ϕ(−q27))/2− 2q7ψ(q2)ψ(q54)(6.38)

= 2qE(q6)E(q18),(6.39)

where in the last step we use (6.15) and (6.18). �

Therefore, b(n) = a(n) − 2d(n). The formulas for b(n) in (6.7) and (6.10) now follow from thosefor a(n) and d(n). Observe also from b(n) = a(n) − 2d(n) that if p is a prime and p ≡ 1 (mod3)then b(p) = 0 if p is represented by (1, 0, 27) and b(p) = 2, otherwise. Hence, these primes cannot berepresented by (1, 0, 27) and (4, 2, 7) at the same time.

While they are not explicitly stated there, the formulas for a(n) and b(n) given by (6.7)–(6.10)can be deduced from Theorem 4.1, Theorem 10.1 and Theorem 10.2 of [19] and and Gauss’ cubicreciprocity law.

7. Representations by the Forms n2 + 5m2 + 5k2 + 5l2 and 5n2 +m2 + k2 + l2

In this section, we give formulas for the number of representations of positive integers by thequaternary forms n2 + 5m2 + 5k2 + 5l2 and 5n2 + m2 + k2 + l2 and also by the restricted formsn+ 5m+ 5k + 5l and 5n+m+ k + l with n,m, k, and l being triangular numbers.

Theorem 7.1.

ϕ(−q)ϕ3(−q5) =(E5(q)

E(q5)+ 4

E5(q2)

E(q10)

)/5−

(qE5(q5)

E(q)− 4q2

E5(q10)

E(q2)

),(7.1)

ϕ3(−q)ϕ(−q5) =(E5(q)

E(q5)+ 4

E5(q2)

E(q10)

)/5− 5

(qE5(q5)

E(q)− 4q2

E5(q10)

E(q2)

),(7.2)

4qψ3(q)ψ(q5) =(E5(q)

E(q5)−E5(q2)

E(q10)

)/5 + 5

(qE5(q5)

E(q)+ q2

E5(q10)

E(q2)

),(7.3)

4q2ψ(q)ψ3(q5) =(E5(q)

E(q5)−E5(q2)

E(q10)

)/5+

(qE5(q5)

E(q)+ q2

E5(q10)

E(q2)

).(7.4)

Furthermore, if a(n), b(n), c(n) and d(n) are defined by

ϕ(q)ϕ3(q5) =: 1 +

∞∑

n=1

a(n)qn, ϕ3(q)ϕ(q5) =: 1 +

∞∑

n=1

b(n)qn,

4qψ3(q)ψ(q5) =:

∞∑

n=1

c(n)qn, 4q2ψ(q)ψ3(q5) =:

∞∑

n=1

d(n)qn,

then

a(n) = (−1)n−1(1 + 5d(−1)g+t)(5 + (−2)g+1)

3

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

1− (−qj)wj+1

1 + qj,(7.5)

b(n) = (−1)n−1(1 + 5d+1(−1)g+t)(5 + (−2)g+1)

3

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

1− (−qj)wj+1

1 + qj,(7.6)

c(n) = (−2)g(−1 + 5d+1(−1)g+t)

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

1− (−qj)wj+1

1 + qj,(7.7)

d(n) = (−2)g(−1 + 5d(−1)g+t)

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

1− (−qj)wj+1

1 + qj,(7.8)

where n has a prime factorization

n = 2g5dr∏

i=1

pvii

s∏

j=1

qwj

j ,

with pi ≡ ±1 (mod5) and qi ≡ ±2 (mod5), qi is odd and t is the number of odd prime divisors of n,counting multiplicities, that are congruent to ±2 (mod5).

Proof. Our proof employs the following well known Lambert series identities of Ramanujan

E5(q)

E(q5)= 1− 5

∞∑

n=1

(n5

) nqn

1− qn,(7.9)

qE5(q5)

E(q)=

∞∑

n=1

(n5

) qn

(1 − qn)2.(7.10)

For the history of these and many related identities see [1], [6, pp. 249–263]. We also use the thetafunction identities [6, p. 262, Entry 10]

(7.11) ϕ2(q)− ϕ2(q5) = 4qf(q, q9)f(q3, q7)

and

(7.12) ψ2(q)− qψ2(q5) = f(q2, q3)f(q, q4).

By multiplying both sides of (7.11) with ϕ3(q5)/ϕ(q), we find that

(7.13) ϕ(q)ϕ3(q5)−ϕ5(q5)

ϕ(q)= 4q

E5(−q5)

E(−q).

From (7.13), we deduce that

16q2E5(q10)

E(q2)= 16q2

ϕ(q)

ϕ5(q5)

E10(−q5)

E2(−q)

=ϕ(q)

ϕ5(q5)

{ϕ(q)ϕ3(q5)−

ϕ5(q5)

ϕ(q)

}2

= ϕ3(q)ϕ(q5)− 2ϕ(q)ϕ3(q5) +ϕ5(q5)

ϕ(q).(7.14)

Using the imaginary transformation on (7.13) and (7.14), we obtain, respectively, that

(7.15) 5ϕ3(q)ϕ(q5)−ϕ5(q)

ϕ(q5)= 4

E5(−q)

E(−q5)

and

(7.16) 25ϕ(q)ϕ3(q5)− 10ϕ3(q)ϕ(q5) +ϕ5(q)

ϕ(q5)= 16

E5(q2)

E(q10).

22 ALEXANDER BERKOVICH AND HAMZA YESILYURT

By multiplying both sides of (7.12) with ψ3(q5)/ψ(q), we find that

(7.17) ψ(q)ψ3(q5)− qψ5(q5)

ψ(q)=E5(q10)

E(q2).

From (7.17), we also find that

E5(q5)

E(q)=

ψ(q)

ψ5(q5)

E10(q10)

E2(q2)

=ψ(q)

ψ5(q5)

{ψ(q)ψ3(q5)− q

ψ5(q5)

ψ(q)

}2

= ψ3(q)ψ(q5)− 2qψ(q)ψ3(q5) + q2ψ5(q5)

ψ(q).(7.18)

Using the imaginary transformation on (7.17) and (7.18), we obtain, respectively, that

(7.19) −5qψ3(q)ψ(q5) +ψ5(q)

ψ(q5)=E5(q2)

E(q10)

and

(7.20) 25q2ψ(q)ψ3(q5)− 10qψ3(q)ψ(q5) +ψ5(q)

ψ(q5)=E5(q)

E(q5).

Using (7.13)–(7.20), we easily derive (7.1)–(7.4).Next, we sketch a proof of (7.5). We omit the proofs of (7.6)–(7.8) since their proofs are similar to

that of (7.5). For convenience, [qn]V (q) will denote the coefficient of qn in the Taylor series expansionof V (q).From (7.9), we have

(7.21)E5(q)

E(q5)= 1− 5

∞∑

n=1

(n5

) nqn

1− qn= 1− 5

∞∑

n=1

(∑

d|n

(d5

)d)qn.

Using the fact that the coefficients are given by the multiplicative function∑

d|n

(d5

)d, we conclude

that for n > 1

(7.22) [qn]E5(q)

E(q5)= −5

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

1− (−qj)wj+1

1 + qj,

where n has a prime factorization

(7.23) n = 5dr∏

i=1

pvii

s∏

j=1

qwj

j ,

with pi ≡ ±1 (mod5) and qi ≡ ±2 (mod5).It is easy to show [11, Thr. 4]

(7.24) [qn]qE5(q5)

E(q)= 5d

r∏

i=1

1− pvi+1i

1− pi

s∏

j=1

(−1)wj1− (−qj)

wj+1

1 + qj,

where n > 0 has a prime factorization

n = 5dr∏

i=1

pvii

s∏

j=1

qwj

j ,

with pi ≡ ±1 (mod5) and qi ≡ ±2 (mod 5). Using (7.22), (7.24) together with (7.1), we arrive at(7.5). �

Theorem 7.1 has the following

Corollary 7.2.

a. [qn](ϕ3(q)ϕ(q5)) > 0, [qn](ψ3(q)ψ(q5)) > 0 for any n ≥ 0,

b. [qn](ϕ(q)ϕ3(q5)) = 0, [qn](ψ(q)ψ3(q5)) > 0 iff n ≡ 2 or 3 (mod5).

Note that our corollary is in agreement with Ramanujan’s observation in [15] where quadratic formx2 + y2 + z2 + 5w2 is listed as universal. It means that this form represents all positive integers.Interested reader may want to check [3] for new results about universal quadratic forms.

Next, we use (7.13)–(7.15) to derive(E5(−q)

E(−q5)+ 4

E5(q2)

E(q10)

)/5

= (5ϕ(q)ϕ3(q5)− ϕ3(q)ϕ(q5))/4(7.25)

=1

4

ϕ2(q5)

ϕ2(q)

{5ϕ3(q)ϕ(q5)−

ϕ5(q)

ϕ(q5)

}(7.26)

=ϕ2(q5)E5(−q)

ϕ2(q)E(−q5)(7.27)

=E5(q2)E7(q10)

E(q)E(q4)E3(q5)E3(q20).(7.28)

Moreover, using (7.9) and (7.10) we obtain(E5(−q)

E(−q5)+ 4

E5(q2)

E(q10)

)/5

= 1 +

∞∑

n=1

(n5

) nqn

1− (−q)n(7.29)

= 1 +

∞∑

n=1

(∑

d|n

(−1)n+dd(d5

))qn.(7.30)

Arguing as above one finds

qE5(−q5)

E(−q)+ 4q2

E5(q10)

E(q2)= q

ϕ2(q)E5(−q5)

ϕ2(q5)E(−q)(7.31)

= qE7(q2)E5(q10)

E3(q)E3(q4)E(q5)E(q20)(7.32)

=∞∑

n=1

(n5

) (−q)n

(1 + (−q)n)2(7.33)

=

∞∑

n=1

(∑

d|n

(−1)n+dd(n/d

5

))qn,(7.34)

and

−(E5(q)

E(q5)−E5(q2)

E(q10)

)/5 = q

ψ2(q5)E5(q2)

ψ2(q)E(q10)(7.35)

= qE2(q)E(q2)E3(q10)

E2(q5)(7.36)

=

∞∑

n=1

(n5

) nqn

1− q2n(7.37)

=

∞∑

n=1

(∑

d|n

d(d5

)γ(n/d)

)qn,(7.38)

24 ALEXANDER BERKOVICH AND HAMZA YESILYURT

where

γ(n) :=

{1 if n is odd,0 if n is even.

Lastly,

qE5(q5)

E(q)+ q2

E5(q10)

E(q2)= q

ψ2(q)E5(q10)

ψ2(q5)E(q2)(7.39)

= qE3(q2)E2(q5)E(q10)

E2(q)(7.40)

=

∞∑

n=1n is odd

(n5

) qn

(1 − qn)2(7.41)

=∞∑

n=1

(∑

d|n

γ(d)(d5

)n/d

)qn.(7.42)

The eta-quotients given by (7.36) and (7.40) are clearly multiplicative. The fact that the first twoeta-quotients given in (7.28) and (7.32) are multiplicative follows from Lemma 5.3. These four multi-plicative eta-quotients are not included in Martin’s list. Lambert series representations of (7.29) and(7.37) are given by Ramanujan [6, p. 249, Entry 8 (i), (ii)]. The identity (7.39)–(7.40) is the identity(6.12) of [4].

8. Outlook

Clearly, this manuscript does not exhaust all potential connections between the Ramanujan iden-tities and quadratic forms. We believe that Ramanujan identities can be employed to find coefficientsof many sextenary forms. For example, in [5] we will show how to use our new identity

7ϕ3(−q)ϕ3(−q7) =− 49(q2E7(q7)

E(q)+ qE3(q)E3(q7)

)

+ 56(7q4

E7(q14)

E(q2)+ q2E3(q2)E3(q14)

)−E7(q)

E(q7)+ 8

E7(q2)

E(q14),

together with two identities of Ramanujan to determine the coefficients of ϕ3(q)ϕ3(q7). There weshall also prove the following intriguing inequalities

[qn](ψ3(q)ψ3(q7)− q

E7(q14)

E(q2)

)≥ 0,

[qn](ϕ3(q)ϕ3(q7) + q7 − q2

E7(q7)

E(q)

)≥ 0.

In the course of this investigation we have determined the coefficients of many multiplicative eta-quotients. Those eta-quotients given by the equations (3.4), (3.5), (4.1), and (6.18) are on Martin’slist [13], while those that are given by (5.1) and (7.28), (7.32), (7.36) and (7.40) are not. The eta-quotients in (7.9) and (7.10) also appear in Martin’s list [13]. In our future publications we plan todetermine the coefficients of all multiplicative eta-products that appear on this list. To this end weproved the following

Theorem 8.1. Suppose the prime factorization of n is

2αr∏

i=1

pui

i

s∏

i=1

qvji

l∏

i=1

Pwi

i

t∏

i=1

Qdi

j ,

where

Qi ≡ 3 (mod 4),

pi ≡ 5 (mod 8),

qi ≡ 1 (mod 8) and 2(qi−1)/4 ≡ 1 (mod qi),

Pi ≡ 1 (mod 8) and 2(Pi−1)/4 ≡ −1 (modPi).

Then

[qn](q

E4(q16)

E(q32)E(q8)

)= δα,0

r∏

i=1

(−1)ui(1 + (−1)ui)/2s∏

i=1

(1 + vi)l∏

i=1

(−1)wj (1 + wj)t∏

i=1

(1 + (−1)di)/2.

We would like to conclude with a following remarkable identity:

qQ̃(2, 0, 7) + Q̃(3, 2, 5)

Q̃(1, 0, 14)− Q̃(3, 2, 5)=E4(q14)E2(q4)

E2(q28)E2(q2),

where

Q̃(a, b, c) :=∞∑

n,m=−∞

qan2+bnm+cm2

.

It is important to observe that neither Q̃(2, 0, 7) + Q̃(3, 2, 5) nor Q̃(1, 0, 14) − Q̃(3, 2, 5) is an eta-quotient. This result along with other similar type identities will be discussed elsewhere.

References

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modular equations of degree 5, Trans. Amer. Math. Soc. 345 (1994), no. 1, 323–345.[18] Li-Chien Shen, On a class of q-series related to Quadratic forms, Bull. Inst. Math. Acad. Sinica 26 (1998), no. 2,

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2, 101–17 1.[20] G. N. Watson, Proof of certain identities in combinatory analysis, J. Indian Math. Soc. 20 (1933), 57–69.[21] K. S. Williams, Some Lambert series expansions of products of theta functions, Ramanujan Journal 3 (1999) ,

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26 ALEXANDER BERKOVICH AND HAMZA YESILYURT

Department of Mathematics, University of Florida, 358 Little Hall, Gainesville, FL 32611, USA